--- title: "Distribution of defectives drawn from ten" subject: "Probability and Statistics" chapter: 2 key: "semester_5/probability_and_statistics:defective-items-three-drawn-distribution" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/defective-items-three-drawn-distribution site: BCT Engineering --- # Distribution of defectives drawn from ten Probability and Statistics · Chapter 2: Discrete Probability Distribution - Asked in: 2068 Jestha (old course), 5 marks - Video: [A package contains 12 capacitors,4 of which are defective.](https://www.youtube.com/watch?v=jkEo1MrYHIM) — Citrus College STEM Center - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/defective-items-three-drawn-distribution - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-2 ## Question Four defective items are mixed with six good items. Three items are drawn at random. Obtain the probability distribution for the number of defective items in the drawn sample. Hence obtain mean and variance for the distribution. ## Short answer Drawing three from 10 items with four defective gives hypergeometric probabilities P(0)=1/6, P(1)=1/2, P(2)=3/10, P(3)=1/30. Mean is 1.2 and variance is 0.56. ## Answer There are 10 items: 4 defective and 6 good. Three are drawn without replacement. Let X = number of defectives in the sample; X is hypergeometric (N = 10, K = 4, n = 3): P(X = x) = C(4, x) · C(6, 3 − x) / C(10, 3), C(10, 3) = 120. ```text x C(4,x)·C(6,3-x) P(x) 0 1 × 20 = 20 20/120 = 1/6 = 0.1667 1 4 × 15 = 60 60/120 = 1/2 = 0.5000 2 6 × 6 = 36 36/120 = 3/10 = 0.3000 3 4 × 1 = 4 4/120 = 1/30 = 0.0333 Sum 1 ``` #### Mean E(X) = Σ x P(x) = 0 + 0.5 + 0.6 + 0.1 = **1.2** (Check: nK/N = 3 × 4/10 = 1.2.) #### Variance E(X²) = 0 + 1(0.5) + 4(0.3) + 9(0.0333) = 0.5 + 1.2 + 0.3 = 2.0 Var(X) = E(X²) − [E(X)]² = 2.0 − 1.44 = **0.56** (Check: n(K/N)(1 − K/N)(N − n)/(N − 1) = 3 × 0.4 × 0.6 × 7/9 = 0.56.) **Answer: distribution as above, mean = 1.2, variance = 0.56** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The number of defectives in the sample of 3 is: A. Binomial with n = 3, p = 0.4 B. Poisson C. Geometric D. Hypergeometric with N = 10, K = 4, n = 3 Answer: D. Draws are without replacement from a finite lot. (Key point: Model) 2. P(X = 1) is: A. C(4,1) C(6,1)/C(10,3) B. C(4,1) C(6,2) / C(10,3) = 0.5 C. C(4,2)C(6,1)/C(10,3) D. 4/10 Answer: B. One defective from 4 and two good from 6, over 120. (Key point: P(X=1)) 3. The mean is: A. n K = 12 B. nK/N = 1.2 C. K/N = 0.4 D. n/N = 0.3 Answer: B. 3 x 4/10 = 1.2. (Key point: Mean) 4. The variance is: A. 2.0 B. 1.2 x 0.6 C. 0.72 D. E(X^2) - 1.44 = 0.56 Answer: D. E(X^2) = 2.0; 2.0 - 1.44 = 0.56, matching the finite population formula. (Key point: Variance) ## Related in Probability and Statistics - [pmf conditions; binomial defective items](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Apmf-conditions-defective-8-items) — Probability and Statistics, chapter 2 - [Sample mean, variance, CV: patients' sleep hours](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Ahospital-patients-sleep-mean-cv) — Probability and Statistics, chapter 1, asked 3× - [Dependent events; defective bolt from machine B](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Abolt-factory-defective-machine-b) — Probability and Statistics, chapter 1 - [Normal: percentage of defective washers](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Awashers-defective-percentage-normal) — Probability and Statistics, chapter 3 - [CLT: sample mean between 67 and 83](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aclt-sample-100-mean-75-variance) — Probability and Statistics, chapter 4, asked 2× - [CLT: sample mean between 75 and 78](https://www.bctengineer.com/question.md?key=semester_5%2Fprobability_and_statistics%3Aclt-sample-100-mean-76-variance) — Probability and Statistics, chapter 4, asked 2× ## Related in other subjects - [RCC slab culvert 1.5 m span: four items](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aslab-culvert-n-four-items) — Estimating and Costing, chapter 6 - [RCC slab culvert (u): four items](https://www.bctengineer.com/question.md?key=semester_7%2Festimating_and_costing%3Aslab-culvert-u-four-items) — Estimating and Costing, chapter 6 - [Random number generation methods](https://www.bctengineer.com/question.md?key=semester_8%2Fsimulation_and_modeling%3Arn-generation-methods) — Simulation and Modeling, chapter 6 - [Random number; problems in generating pseudo-random](https://www.bctengineer.com/question.md?key=semester_8%2Fsimulation_and_modeling%3Arn-problems-pseudo-random) — Simulation and Modeling, chapter 6